Induced charge in image problem

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Homework Help Overview

The discussion revolves around finding the induced charge density in a classic image problem related to electrostatics. Participants are examining the application of potential theory and boundary conditions in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use polar coordinates but expresses uncertainty about their approach. Some participants question the validity of the dipole approximation in this scenario, while others suggest reconsidering the boundary conditions and the use of cylindrical coordinates.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the boundary conditions and potential expressions. There is no explicit consensus yet, but guidance has been offered regarding the formulation of the potential and the coordinate system to use.

Contextual Notes

Participants are grappling with the implications of the potential not going to zero on the plane and the requirement for the entire plane to be an equipotential surface. The original poster's use of an image representation of their work may also limit clarity in the discussion.

Kaguro
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Homework Statement


Find induced charge density in the classic image problem.

Sorry, I uploaded my work in form of an image instead of typing.

I have tried to use polar coordinates to solve this. But i must have done something wrong. Please help me find the mistake.

15512082946811688506627.jpg
 

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I do not believe you can use the dipole approximation here. It is valid for ##r>>d##. Your potential does not go to zero on the plane which is the boundary condition in this problem.
 
kuruman said:
I do not believe you can use the dipole approximation here. It is valid for ##r>>d##. Your potential does not go to zero on the plane which is the boundary condition in this problem.

But my potential does go to 0 on the plane as theta is 90 then and so cos theta is 0.
 
The entire plane is an equipotential therefore the expression must evaluate to zero for any value of angle ##\theta## and must also be zero at the origin. You should express the potential as the sum of the potentials from the point charge and its image. You would be wise to use cylindrical coordinates.
 

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